CSIR NET Mathematics Mock Test - 6 - CSIR NET Mathematics MCQ

SPICE (Simulation Program with Integrated Circuit Emphasis) was introduced in May 1972 by the University of Berkeley, California.

Strong nuclear force is the strongest force. It is present inside the atom responsible for binding the protons and neutrons and also inside the proton and neutron in binding up the quarks.

The series is pqp/pqp/pqp/pqp. Thus, the pattern 'pqp' is repeated.

Rise of mercury indicates increase in atmospheric pressure. As air descends, it warms and contracts, which reduces or prevents the formation of clouds. Because of this effect, areas of high pressure often create clear, dry weather.

Total runs of 57 innings
∴ Average of remaining or 50 innings

Sum of the score of 24 students = 24 x 54

Actual Sum of the scores of 24 students = 24 x 54 + 88 - 64 = 1320

Actual Average = 1320/24 = 55

eq. (1)

(3)

By eq. (1)

By eq (3)

We know



So choice ( 2 ) is answer

⇒ i A is Skew-Hermitian.

The determinant exists only for square matrix

∴ m = n.

The inverse of a matrix is unique

The th term

is does not converges to zero is divergent.

is a bounded sequence

is converges to zero
is bounded sequence and conver ges to zero converges to zero.

and the radius of convergence

Option B is correct answer.

If f (0) < f (1) then f ([0, 1]) must be equal to [f (0), f (1)]

Hence option B is correct.

(It's not true that any cyclic subgroup of a group is normal. You can see this since for example the symmetric group has a cyclic subgroup of order 2. It is not normal because 
By Lagrange's theorem, the non-trivial proper subgroups have order 3 or 7.
As you have correctly identified, from Sylow's theorems, G has a unique subgroup N of order 7. It must be normal, since for any prime number P the Sylow P -subgroups of a group form a single conjugacy class of subgroups.
Suppose (for contradiction) that it also has a normal subgroup K of order 3. Then
(By Lagrange's theorem, the order of divides 3 and 7, so is 1 ). Their product NK is thus the whole group G, since it has order (To see this consider the map So any and commute. (Consider an element of the form
We would thus have G is isomorphic to the direct product of N and K. In particular G would be abelian, which is a contradiction.
So G has no normal subgroup of order 3, and by Sylow's theorems has only one of order 7. Hence option 2 is correct.

Given

Taking Laplace transform of both sides, we get

or

or

or

On inversion, we obtain

or

Given differential equation is

S.F. is

A. is

is sin

Now we find I.

Now

So general solution is

Let sequence 〈 x_n 〉 is a Cauchy sequence, for

we have

The sequence is bounded.

Solution : Let,

ab is an upper bound for C

C is bounded above

By completeness axiom, C has a sup C = c
(say)
c is a cup C

Now, we have to prove
Let
...........(1)

Choose

Then, we have

................(2)

By (1) and (2)

Since φ and ψ are simple functions, it can be represented by canonical representation

Let , then they form a disjoint collection of measurable sets and can be represented as,

Take A = −I= −I. Then A³ = −1 but A does not have distinct eigenvalues.
The minimal polynomial of a matrix A may be defined as the polynomial of smallest degree that is satisfied by A and has highest coefficient equal to 1.

Ans (D) n = 3000 will ensure | p – θ | ≤ 0.02 with probability at least 0.95.